Simplify cube root of 1+ square root of -100-3 square root of 16-2 square root of -49

$\sqrt[3]{1} + \sqrt{- 100} - 3 \sqrt{16} - 2 \sqrt{- 49}$

Step 1

Simplify each term.

Any root of $1$ is $1$ .

$1 + \sqrt{- 100} - 3 \sqrt{16} - 2 \sqrt{- 49}$

Rewrite $- 100$ as $- 1 (100)$ .

$1 + \sqrt{- 1 (100)} - 3 \sqrt{16} - 2 \sqrt{- 49}$

Rewrite $\sqrt{- 1 (100)}$ as $\sqrt{- 1} \cdot \sqrt{100}$ .

$1 + \sqrt{- 1} \cdot \sqrt{100} - 3 \sqrt{16} - 2 \sqrt{- 49}$

Rewrite $\sqrt{- 1}$ as $i$ .

$1 + i \cdot \sqrt{100} - 3 \sqrt{16} - 2 \sqrt{- 49}$

Rewrite $100$ as $10^{2}$ .

$1 + i \cdot \sqrt{10^{2}} - 3 \sqrt{16} - 2 \sqrt{- 49}$

Pull terms out from under the radical, assuming positive real numbers.

$1 + i \cdot 10 - 3 \sqrt{16} - 2 \sqrt{- 49}$

Move $10$ to the left of $i$ .

$1 + 10 \cdot i - 3 \sqrt{16} - 2 \sqrt{- 49}$

Rewrite $16$ as $4^{2}$ .

$1 + 10 i - 3 \sqrt{4^{2}} - 2 \sqrt{- 49}$

Pull terms out from under the radical, assuming positive real numbers.

$1 + 10 i - 3 \cdot 4 - 2 \sqrt{- 49}$

Multiply $- 3$ by $4$ .

$1 + 10 i - 12 - 2 \sqrt{- 49}$

Rewrite $- 49$ as $- 1 (49)$ .

$1 + 10 i - 12 - 2 \sqrt{- 1 (49)}$

Rewrite $\sqrt{- 1 (49)}$ as $\sqrt{- 1} \cdot \sqrt{49}$ .

$1 + 10 i - 12 - 2 (\sqrt{- 1} \cdot \sqrt{49})$

Rewrite $\sqrt{- 1}$ as $i$ .

$1 + 10 i - 12 - 2 (i \cdot \sqrt{49})$

Rewrite $49$ as $7^{2}$ .

$1 + 10 i - 12 - 2 (i \cdot \sqrt{7^{2}})$

Pull terms out from under the radical, assuming positive real numbers.

$1 + 10 i - 12 - 2 (i \cdot 7)$

Move $7$ to the left of $i$ .

$1 + 10 i - 12 - 2 (7 \cdot i)$

Multiply $7$ by $- 2$ .

$1 + 10 i - 12 - 14 i$

Step 2

Simplify by adding terms.

Subtract $12$ from $1$ .

$- 11 + 10 i - 14 i$

Subtract $14 i$ from $10 i$ .

$- 11 - 4 i$